Question

如图,已知AB⊥平面ACD,DE⊥平面ACD,△ACD是正三角形,且AD=DE=2AB,F是CD的中点.

(1)求证:平面CBE⊥平面CDE;

(2)求二面角F-BE-C的大小.

Answer 1

(1)证明:∵DE⊥平面ACD,?

∴平面CDE⊥平面ACD.?

又∵AF⊥CD,?

∴AF⊥平面CDE.?

取CE的中点N,连结FN,BN,如图所示.?

∴FN DE=AB.                                                                                               ?

∴四边形AFNB为平行四边形.                                                                                ?

∴BN∥AF.

∴BN⊥平面CDE.?

∴平面CBE⊥平面CDE.                                                                                         .?

(2)解:建立如图所示的空间直角坐标系.?

设AB=1,则B(0,,1),F(0,0,0),C(-1,0,0),=(0,3,), =(1,0,2),=(1,3,1), =(2,0,2).                                                                                                         7分?

设平面FEB的法向量为n=(x,y,z),平面BCE的法向量为M=(p,q,r).?

则                                                                                                       ?

即?

令y=1,则n=(2,1,-).?

即.?

令r=1,则M=(-1,0,1).                                                                                        ?

∴cos〈m,n〉===-.                                            ?

∴二面角F-BE-C的大小为π-arccos.                                                              ?